Show that 9ⁿ ⁺ ¹ - 8n - 9 is divisible by 64, whenever n is a positive integer

Solution:

To show that 9ⁿ ⁺ ¹ - 8n - 9 is divisible by 64, we need to prove that 9ⁿ ⁺ ¹ - 8n - 9 = 64k where k is some natural number.

By binomial theorem,

(1 + a)ᵐ = ᵐC₀ + ᵐC₁ a + ᵐC₂ a² + .... + ᵐCₘ aᵐ

For a = 8 and m = n + 1, we obtain

(1 + 8)ⁿ ⁺ ¹ = ⁿ ⁺ ¹C₀ + ⁿ ⁺ ¹C₁ a + ⁿ ⁺ ¹C₂ a² + .... + ⁿ ⁺ ¹Cₙ ₊ ₁ aⁿ ⁺ ¹

Hence,

9ⁿ ⁺ ¹ = 1+ (n + 1) (8) + ⁿ ⁺ ¹C₂ (8)² + .... + ⁿ ⁺ ¹Cₙ ₊ ₁ (8)ⁿ⁺¹

9ⁿ ⁺ ¹ = 9 + 8n + (8)² [ⁿ ⁺ ¹C₂ + .... + ⁿ ⁺ ¹Cₙ ₊ ₁ (8)ⁿ⁻¹]

Therefore, 9ⁿ ⁺ ¹ - 8n - 9 = 64k, where k = [ⁿ ⁺ ¹C₂ + .... + ⁿ ⁺ ¹Cₙ ₊ ₁ (8)ⁿ⁻¹] is a natural number.

Thus, 9ⁿ ⁺ ¹ - 8n - 9 is divisible by 64, whenever n is a positive integer

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.1 Question 13

Show that 9ⁿ ⁺ ¹ - 8n - 9 is divisible by 64, whenever n is a positive integer

Summary:

Using the binomial theorem, we have proved that 9ⁿ ⁺ ¹ - 8n - 9 is divisible by 64, whenever n is a positive integer